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a,b,c are three different numbers.None of the numbers equals the

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a,b,c are three different numbers.None of the numbers equals the  [#permalink]

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New post Updated on: 19 Apr 2017, 13:36
6
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

69% (02:11) correct 31% (02:24) wrong based on 64 sessions

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a,b,c are three different numbers.None of the numbers equals the average of the other two.
If \(\frac{x}{{a+b-2c}}\) = \(\frac{y}{{b+c-2a}}\) = \(\frac{z}{{c+a-2b}}\) , then what is the value of \(x+y+z\) ?

A)-1
B)5
C)3
D)1
E)0


Source-> NOVA.

_________________

Originally posted by stonecold on 19 Apr 2017, 13:32.
Last edited by Bunuel on 19 Apr 2017, 13:36, edited 1 time in total.
Renamed the topic.
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Re: a,b,c are three different numbers.None of the numbers equals the  [#permalink]

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New post 19 Apr 2017, 13:36
3
1
Let

x/(a+b-2c) = y/(b+c-2a) = z/(c+a-2b) = k

which means

x = (a+b-2c)k
y = (b+c-2a)k
z = (c+a-2b)k

x+y+z = k*[a+b-2c+b+c-2a+c+a-2b] = k*0 = 0

Answer is E)0
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a,b,c are three different numbers.None of the numbers equals the  [#permalink]

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New post 03 Mar 2019, 19:18
stonecold wrote:
a,b,c are three different numbers.None of the numbers equals the average of the other two.
If \(\frac{x}{{a+b-2c}}\) = \(\frac{y}{{b+c-2a}}\) = \(\frac{z}{{c+a-2b}}\) , then what is the value of \(x+y+z\) ?

A)-1
B)5
C)3
D)1
E)0


Source-> NOVA.


Hello stonecold !

Could you please provide the explanation?

Kind regards!
GMAT Club Bot
a,b,c are three different numbers.None of the numbers equals the   [#permalink] 03 Mar 2019, 19:18
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a,b,c are three different numbers.None of the numbers equals the

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